Question

2 x²-9 1 x² + 3x 1 2 x² - 6x +9 + 3​

Answer 1

Ответ:

1

Объяснение:

$(\frac{2}{x^2-9} - \frac{1}{x^2+3x}) : \frac{1}{x^2 - 6x + 9} + \frac{3}{x} = \frac{2x-(x-3)}{x*(x-3)*(x+3)} * (x-3)^2 + \frac{3}{x} = \frac{x+3}{x*(x-3)*(x+3)} * (x-3)^2 + \frac{3}{x} = \frac{1}{x*(x-3)} * (x-3)^2 + \frac{3}{x} = \frac{x - 3}{x} + \frac{3}{x} = \frac{x}{x} = 1$